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Sorry made a few mistakes in my original post.
One being that I used the @name expression and not the node( ) function to get the name attribute of the result-generating node. Ive corrected the mistakes in my post below.
Rahil wrote:
Re: [xsl] backtracking to find all parents till root[again]
Subject: Re: [xsl] backtracking to find all parents till root[again] From: Rahil <qamar_rahil@xxxxxxxxxxx> Date: Tue, 24 May 2005 11:52:05 +0100 |
Sorry made a few mistakes in my original post.
One being that I used the @name expression and not the node( ) function to get the name attribute of the result-generating node. Ive corrected the mistakes in my post below.
Apologies Rahil
Rahil wrote:
Hi
How do I backtrack to the root element from the result-generating node? Hence if my given structure is of the form:
<Top> <SubConcepts> <SubConcept id="990" name="Level1"> <Child ref="567">Child1</Child> <Value ref="456">hasFeature BrokenBolt</Value> </SubConcept> <SubConcept id="456" name="BrokenBolt"> <Child ref="345">Child2</Child> <Value ref="123">hasProperty NextTime</Value> </SubConcept> </SubConcepts> </Top>
I find my result in the <Value ref="123"> node with the contained value 'Time'. I can find the parent of this with the @name expression to get 'BrokenBolt' as the control is at the <SubConcept id="456"> level. I'd like to iterate backwards using the 'id'='ref' match so I get a result of the order:
Match found: Next Parent: BrokenBolt Parent: Level1 Parent: SubConcepts
Thanks Rahil
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